There is a real-valued random variable $R$. Define a finite set of random variables ("data points") $$X_i = R + Z_i \; \text{for } i\in\{1,\ldots,n\},$$ where $Z_i$ are identically and independently distributed, mean-zero, and independent of $R$. The prior distributions of $R$ and the $Z_i$ are given; the support of each is the real line.

Define the conditional, or posterior, expectation of $R$ given particular realizations of the data $$\hat{R}(x_1,\ldots,x_n)= \mathbb{E}[R \mid X_1=x_1,\ldots,X_n=x_n]. $$ (This is the expectation of $R$ assessed by someone who sees the data points but not $R$ itself, and is a measurable function $\mathbb{R}^n \to \mathbb{R}$.) The question is: how much can an adversary with a given amount of manipulation power over a single data point move this estimate?

More precisely, fix a number $\Delta$. I am looking for a bound, which is useful as $n$ grows large, on $$M_n(\Delta)=\mathbb{E}[\hat{R}(X_1+\Delta,X_2,\ldots,X_n) - \hat{R}(X_1,X_2,\ldots,X_n)].$$ This expectation is an integral over all uncertainty in the model (i.e. in $R$ and the $X_i$), though I believe it should be possible to give a good bound even conditional on $R$.

We can take all random variables to be square-integrable if necessary, and make any other convenient assumptions.

A conjecture is that the manipulability is small in the sense that $M(\Delta) \to 0 $ as $n \to \infty$ and indeed $M_n'(\Delta) \to 0$ as well.

The conclusion may seem obvious because the posterior distribution of $R$ conditional on the data $X_1,\ldots,X_n$ converges to a point mass whose location is independent of the realization of $X_1$. But this does not readily imply a bound on the $L^1$ norm of the difference between $\hat{R}(X_1+\Delta,\ldots,X_n)$ and $R$, or the difference between $\hat{R}(X_1+\Delta,\ldots,X_n)$ and the unmanipulated estimate $\hat{R}(X_1,\ldots,X_n)$. It could be that the manipulation has a slowly decaying probability of achieving very large deviations in the estimate, so that it messes up the expectation.

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    $\begingroup$ Do we know anything about the dependence between $\theta$ and the $\epsilon_i$? $\endgroup$ Aug 4, 2019 at 19:25
  • $\begingroup$ Independent of $\theta$, thanks! $\endgroup$
    – Ben Golub
    Aug 4, 2019 at 19:29
  • $\begingroup$ So then, doesn't $\hat{\theta}$ just equal $(X_1+\dots+X_n)/n - \mu$ where $\mu = \mathbb{E}[\epsilon_i]$? Then the quantity you're asking about is exactly equal to $\delta/n$. $\endgroup$ Aug 4, 2019 at 19:30
  • $\begingroup$ In general, the posterior mean won't satisfy that formula. Among other things, if you have a very strong prior that $\theta$ is near some value, then you'll have to adjust your estimate in that direction (this is so even when all rv's are Gaussian). More generally such adjustments will take a complicated form. $\endgroup$
    – Ben Golub
    Aug 4, 2019 at 19:34
  • $\begingroup$ Do you also assume that the $\epsilon$’s have distributions with mean 0, median 0, or symmetry about the origin? $\endgroup$
    – Matt F.
    Aug 5, 2019 at 4:12

1 Answer 1


There is no bound independent of $R$.

In what follows, I use my proposed notation, with $Y$ instead of $R$. Take \begin{align} Z &\sim N(0,1) \\ Y &\sim \text{even mix of } N(b,1) \text{ and } N(-b,1) \\ X &\sim \text{even mix of } N(b,\sqrt{2}) \text{ and }N(-b,\sqrt{2}) \end{align} So \begin{align} P(Z=z) &= \frac{1}{\sqrt{2\pi}\ \ }\,e^{-z^2/2} \\ P(Y=y) &= \frac{1}{2\sqrt{2\pi}}\left(e^{-(y-b)^2/2} + e^{-(y+b)^2/2}\right)\\ P(X=x) &= \frac{1}{\ 4\sqrt{\pi}\ }\left(e^{-(x-b)^2/4} + e^{-(x+b)^2/4}\right) \end{align} Suppose we have a single observation, namely $x$. Then \begin{align} P(Y'=y|X=x) &= \frac{P(x|y)P(y)}{P(x)}\\ &= \frac{\frac{1}{\sqrt{2\pi}\ \ }\,e^{-(x-y)^2/2}\frac{1}{2\sqrt{2\pi}}\left(e^{-(y-b)^2/2} + e^{-(y+b)^2/2}\right)} {\frac{1}{\ 4\sqrt{\pi}\ }\left(e^{-(x-b)^2/4} + e^{-(x+b)^2/4}\right)}\\ &= \frac{e^{-(x^2+b^2)/2}\left(e^{-y^2+by+xy} + e^{-y^2-by+xy}\right)} {\sqrt{\pi}\left(e^{-(x-b)^2/4} + e^{-(x+b)^2/4}\right)} \\ \\ E[Y'|X=x] &= \int_{y=-\infty}^\infty y\,P(Y'=y|X=x)\,dy\\ &= \frac{e^{-(x^2+b^2)/2}\left((x+b)e^{(x+b)^2/4} + (x-b)e^{(x-b)^2/4}\right)} {2\left(e^{-(x-b)^2/4} + e^{-(x+b)^2/4}\right)} \\ &= \frac{(x+b)e^{bx/2} + (x-b)e^{-bx/2}} {2\left(e^{bx/2} + e^{-bx/2}\right)} \\ \\ \frac{dE[Y'|X=x]}{dx}{\Large|}_{x=0} &= \frac{b^2+2}{4} \end{align} So the expectation of $Y'$ can be made to depend on $x$ with arbitrarily large sensitivity. Any bound on this sensitivity would likely be of the order of the variance of $Y$.

  • $\begingroup$ Where is $n$ in this answer? It's obvious you can't give a good bound fixing $n=1$ or 2. The question is whether the manipulation power over one data point grows small as one accumulates many other unmanipulated data points. See comments at the end the question. $\endgroup$
    – Ben Golub
    Aug 5, 2019 at 17:37
  • $\begingroup$ If there are $n$ observations, can't we just repeat the analysis where $Y'$ incorporates the first $n-1$, and then $Y''$ incorporates the one new observation? Meanwhile I expect a direct formula for the sensitivity would be roughly $b^2/4n$. $\endgroup$
    – Matt F.
    Aug 5, 2019 at 17:48
  • $\begingroup$ So in your example, the conjecture that sensitivity decays with n would be satisfied, for a given distribution of R and upper bound on the manipulation. But the question is whether this holds for all distributions. $\endgroup$
    – Ben Golub
    Aug 5, 2019 at 19:41

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